{"id":7036,"date":"2021-10-22T15:45:17","date_gmt":"2021-10-22T10:15:17","guid":{"rendered":"https:\/\/mathemerize.com\/?p=7036"},"modified":"2021-10-25T01:37:09","modified_gmt":"2021-10-24T20:07:09","slug":"a-biased-coin-with-probability-p-0-p-1-of-heads-is-tossed-until-a-head-appears-for-the-first-time-if-the-probability-that-the-number-of-tosses-required-is-even-is-2-5-then-p-is-equal-to","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/a-biased-coin-with-probability-p-0-p-1-of-heads-is-tossed-until-a-head-appears-for-the-first-time-if-the-probability-that-the-number-of-tosses-required-is-even-is-2-5-then-p-is-equal-to\/","title":{"rendered":"A biased coin with probability p, 0 < p < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even, is 2\/5, then p is equal to"},"content":{"rendered":"

Solution :<\/h2>\n

Let the probability of getting a head be p and not getting a head be q.<\/p>\n

Since, head appears first time in an even throw 2 or 4 or 6.<\/p>\n

\\(\\therefore\\)\u00a0 \u00a0\\(2\\over 5\\) = qp + \\(q^3\\)p + \\(q^5\\)p + ……<\/p>\n

\\(\\implies\\)\u00a0 \\(2\\over 5\\) = \\(qp\\over {1 – q^2}\\)<\/p>\n

Since q = 1- p<\/p>\n

\\(\\implies\\)\u00a0 \\(2\\over 5\\) = \\((1 – p)p\\over {1 – (1 – p)^2}\\)<\/p>\n

\\(\\implies\\)\u00a0 \\(2\\over 5\\) = \\((1 – p)\\over {2 – p}\\)<\/p>\n

\\(\\implies\\)\u00a0 4 – 2p = 5 – 5p<\/p>\n

\\(\\implies\\)\u00a0 p =\\(1\\over 3\\)<\/p>\n

\n

Similar Questions<\/h3>\n

The probability of India winning a test match against the west indies is 1\/2 assuming independence from match to match. The probability that in a match series India\u2019s second win occurs at the third test is<\/a><\/p>\n

A fair die is tossed eight times. The probability that a third six is observed on the eight throw, is<\/a><\/p>\n

If A and B are two mutually exclusive events, then<\/a><\/p>\n

A and B play a game, where each is asked to select a number from 1 to 25. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial is<\/a><\/p>\n

A problem in mathematics is given to three students A, B, C and their respective probability of solving the problem is \\(1\\over 2\\), \\(1\\over 3\\) and \\(1\\over 4\\). Probability that the problem is solved is<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Let the probability of getting a head be p and not getting a head be q. Since, head appears first time in an even throw 2 or 4 or 6. \\(\\therefore\\)\u00a0 \u00a0\\(2\\over 5\\) = qp + \\(q^3\\)p + \\(q^5\\)p + …… \\(\\implies\\)\u00a0 \\(2\\over 5\\) = \\(qp\\over {1 – q^2}\\) Since q = 1- …<\/p>\n

A biased coin with probability p, 0 < p < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even, is 2\/5, then p is equal to<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,44],"tags":[],"yoast_head":"\nA biased coin with probability p, 0 < p < 1, of heads is tossed until a head appears for the first time. 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Since, head appears first time in an even throw 2 or 4 or 6. (therefore)\u00a0 \u00a0(2over 5) = qp + (q^3)p + (q^5)p + …… (implies)\u00a0 (2over 5) = (qpover {1 – q^2}) Since q = 1- … A biased coin with probability p, 0 < p < 1, of heads is tossed until a head appears for the first time. 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